Problem: The equation of hyperbola $H$ is $\dfrac {(y-7)^{2}}{81}-\dfrac{x^2}{36} = 1$. What are the asymptotes?
Answer: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y-7)^{2}}{81} = 1 + \dfrac{x^2}{36}$ Multiply both sides of the equation by $81$ $(y-7)^{2} = { 81 + \dfrac{ x^2 \cdot 81 }{36}}$ Take the square root of both sides. $\sqrt{(y-7)^{2}} = \pm \sqrt { 81 + \dfrac{ x^2 \cdot 81 }{36}}$ $ y - 7 = \pm \sqrt { 81 + \dfrac{ x^2 \cdot 81 }{36}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y - 7 \approx \pm \sqrt {\dfrac{ x^2 \cdot 81 }{36}}$ $y - 7 \approx \pm \left(\dfrac{9 \cdot (x)}{6}\right)$ Add $7$ to both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{3}{2}(x)+ 7$